-9z^2+14z+24=0

Solution for -9z^2+14z+24=0 equation:

-9z^2+14z+24=0

a = -9; b = 14; c = +24;
Δ = b2-4ac
Δ = 142-4·(-9)·24
Δ = 1060
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1060}=\sqrt{4*265}=\sqrt{4}*\sqrt{265}=2\sqrt{265}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-2\sqrt{265}}{2*-9}=\frac{-14-2\sqrt{265}}{-18}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+2\sqrt{265}}{2*-9}=\frac{-14+2\sqrt{265}}{-18}
$